CHAPTER 3 BIOCHEMISTRY - maloyscience Century Arts€¦ · BIOCHEMISTRY 51 Liquid movement up a stem WATER DYE 1. Describe the structure of a water molecule. 2. ... The chemistry

BIOCHEMISTRY

FOCUS CONCEPT: Matter, Energy, andOrganization

As you read this chapter, notice how function depends on structure in eachof the compounds you examine.

The body of this jellyfish, Pseudorhiza haeckeli, is almost 99 percent water.

3-1 Water

3-2 Carbon Compounds

3-3 Molecules of Life

48

CHAPTER 3

Copyright © by Holt, Rinehart and Winston. All rights reserved.


WA T E RCompare the body of the jellyfish shown on the opposite page

with your own body. The jellyfish will die if it is removed from

its water environment. You can live in the driest parts of Earth.

Jellyfish and humans seem utterly unlike each other, yet the

bodies of both are made of cells filled with water. The chemical

reactions of all living things take place in an aqueous environ-

ment. Water has several unique properties that make it one of

the most important compounds found in living things.

POLARITYMany of water’s biological functions stem from its chemical struc-ture. Recall that in the water molecule, H2O, the hydrogen and oxy-gen atoms share electrons to form covalent bonds. However, theseatoms do not share the electrons equally. An oxygen atom haseight protons in its nucleus and therefore eight positive charges toattract electrons, whereas a hydrogen atom has only one protonand therefore one positive charge. With its greater positive charge,the nucleus of the oxygen atom pulls the shared electrons towardits nucleus and away from the nucleus of the hydrogen atom. As aresult, the electrical charge is unevenly distributed, as shown inthe models of a water molecule shown in Figure 3-1.

Notice too in Figure 3-1 that the three atoms in a water moleculeare not arranged in a straight line as you might expect. Rather, thetwo hydrogen atoms bond with the single oxygen atom at an angle.Although the total electrical charge on a water molecule is neutral,the region of the molecule where the oxygen atom is located has a

49B I O C H E M I S T R Y

HH

OH

O

H

(a) Electron-energy-level model (b) Structural formula (c) Space-filling model

The oxygen region of the water mol-ecule is weakly negative, and the hydro-gen regions are weakly positive. Noticethe three very different ways to representwater, H2O. You are familiar with the electron-energy-level model (a) fromChapter 2. The structural formula (b) iscompact and easy to understand. Thespace-filling model (c) shows the three-dimensional structure of a molecule.

FIGURE 3-1

▲

Describe the structure of a water molecule.

●

Explain how water’s polar natureaffects its ability to dissolve

substances.

■

List two of water’s properties that result from

hydrogen bonding.

O B J E C T I V E S

SECTION

3-1

� � � �

�

�


slightly negative charge, while the regions of the molecule whereeach of the two hydrogen atoms are located have a slightly positivecharge. Because of this uneven pattern of charge, water is called apolar compound.

It is this polar nature that makes water very effective in dissolv-ing many other substances. Water dissolves other polar substances,including sugars and some proteins, as well as ionic compounds,such as sodium chloride, NaCl. An ionic compound mixed withwater tends to dissociate into ions. This is illustrated in Figure 3-2.This breaking up of an ionic compound frees ions to participate inmany biological reactions. In your body, both sodium ions and chlo-ride ions are essential to functions like muscle contraction andtransmission of impulses in the nervous system. In fact, dissolved,dissociated ions are present in all of the aqueous solutions found inliving things. Their concentration is critical to the normal operationof the many systems of your body.

HYDROGEN BONDING

The polar nature of water also causes water molecules to beattracted to one another. The type of attraction that holds twowater molecules together is called a hydrogen bond. As shown inFigure 3-3, a positive region of one water molecule is attracted tothe negative region of another water molecule. Thus, a hydrogenbond tends to form between a hydrogen atom in one molecule andthe region of negative charge on another molecule. A hydrogenbond is a weak bond that can be easily broken. Even so, the hydro-gen bonds in water exert a significant attractive force, causingwater to cling to itself and to other substances.

C H A P T E R 350

The positive end of a water moleculeattracts the negative end of an ioniccompound, such as the Cl� portion ofNaCl. Similarly, the negative end of thewater molecule attracts the positiveend of the compound—the Na� por-tion of NaCl. As a result, NaCl breaksapart, or dissociates, in water.

FIGURE 3-2

The dotted lines in this figure representhydrogen bonds. A hydrogen bond is a weak force of attraction between a hydrogen atom in one molecule and a negatively charged atom in a secondmolecule.

FIGURE 3-3

�

��

�


Cohesion andAdhesionAn attractive force betweenparticles of the same kind isknown as cohesion. You cansee cohesion at work when you observe the surface ten-sion of water. Cohesive forcesresulting from water’s hydro-gen bonding are strong enoughto cause water to act as if it hasa thin “skin” on its surface.This is why water appears tobulge from the sides of a glassfilled to the brim.

Adhesion is the attractiveforce between unlike sustances.Together, adhesion and cohe-sion enable water molecules tomove upward through narrowtubes against the force of gravity. This property of water is known ascapillarity (KAP-uh-LER-it-ee). You have seen capillarity at work if youhave observed the flow of water into a flower through its stem, suchas is shown in Figure 3-4.

Temperature ModerationWater must gain or lose a relatively large amount of energy for its temperature to change. When water is heated, most of the ther-mal energy that the water initially absorbs breaks the hydrogenbonds between the molecules. Only after these bonds have beenbroken does the thermal energy increase the motion of the mol-ecules and raise the temperature of the water. You read in Chap-ter 1 that all organisms must maintain homeostasis to live. Inorganisms, water’s ability to absorb large amounts of energy helpskeep cells at an even temperature despite temperature changes inthe environment.


Liquid movement up a stem

WATER DYE

1. Describe the structure of a water molecule.

2. How do molecules of a polar compound differfrom those of a nonpolar compound?

3. What happens when ionic compounds are mixedwith water?

4. What are two properties of water that resultfrom water’s tendency to form hydrogen bonds?

5. What is capillarity?

6. CRITICAL THINKING Most automobiles havewater-cooled engines. What must be true about a solution that can replace water in the coolingsystem, such as antifreeze?

SECTION 3-1 REVIEW

Because of strong cohesive and adhe-sive forces, water can travel upwardfrom the roots of flowers. In the floweron the right, the water, which has beendyed blue, has moved up through thestem to the flower’s petals.

FIGURE 3-4

TOPIC: Hydrogen bondingGO TO: www.scilinks.orgKEYWORD: HM051


C A R B O N C O M P O U N D SAll of the many compounds discovered can be classified in

two broad categories: organic compounds and inorganic

compounds. Organic compounds contain carbon atoms that

are covalently bonded to other carbon atoms and to other

elements as well—typically hydrogen, oxygen, and nitrogen.

The chemistry of carbon is the chemistry of life.

CARBON BONDINGA carbon atom has four electrons in its outermost energy level.Remember from Chapter 2 that most atoms become stable whentheir outermost energy level contains eight electrons. A carbonatom therefore readily forms four covalent bonds with other ele-ments. Unlike other elements, however, carbon also readily bondswith other carbon atoms, forming straight chains, branched chains,or rings, as shown in Figure 3-5. This tendency of carbon to bondwith itself results in an enormous variety of organic compounds.

In the symbolic shorthand of chemistry, each line shown inFigure 3-5 represents a covalent bond formed when two atomsshare a pair of electrons. A bond formed when two atoms share apair of electrons is called a single bond. Carbon can also share twoor even three pairs of electrons with another atom. Figure 3-6ashows a model for an organic compound in which six carbon atomshave formed a ring. Notice that each carbon atom forms four cova-lent bonds: a single bond with another carbon atom, a single bondwith a hydrogen atom, and a double bond with a second carbonatom. In a double bond—represented by two parallel lines—atomsshare two pairs of electrons. A triple bond, the sharing of threepairs of electrons, is shown in Figure 3-6b.

C H A P T E R 352

Straight chain

H C

H

H

H

H

H

H

H

H

C C C H

H

H

H

C

H

H

H

H

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

C

C

CC C

H H

H

Branched chain Ring

Carbon can bond in a number of waysto produce molecules of very differentshapes, including straight chains,branched chains, and rings. Thesestructures form the backbone of manydifferent kinds of organic molecules.

FIGURE 3-5

SECTION

3-2

▲

Define organic compound and name three elements often found

in organic compounds.

●

Explain why carbon forms so many different compounds.

■

Define functional group and explain its significance.

◆

Compare a condensation reaction with hydrolysis.

O B J E C T I V E S


FUNCTIONAL GROUPSIn most organic compounds, clusters of atoms, called functionalgroups, influence the properties of the molecules they compose. Thefunctional group is the structural building block that determines thecharacteristics of the compound. One functional group important toliving things, the hydroxyl group, —OH, is shown in Figure 3-6c.

An alcohol is an organic compound with a hydroxyl groupattached to one of its carbon atoms. Locate the hydroxyl group inthe alcohol shown in Figure 3-6c. The hydroxyl group makes analcohol a polar molecule. Thus, alcohols have some propertiessimilar to water, including the ability to form hydrogen bonds. Thealcohol illustrated in Figure 3-6c is ethanol, which is found in alco-holic beverages. Ethanol causes cell death in the liver and brain ofhumans. The alcohol methanol, also called wood alcohol, cancause blindness or even death when consumed. Some alcohols,however, are needed by organisms to carry out their life processes.Humans, for example, need the alcohol glycerol to assemble cer-tain molecules necessary for life.

LARGE CARBON MOLECULESIn many carbon compounds, the molecules are built up from smaller,simpler molecules known as monomers, such as the ones shown inFigure 3-7. As you can also see in Figure 3-7, monomers can bond to one another to form complex molecules known as polymers. Apolymer consists of repeated, linked units. The units may be identi-cal or structurally related to each other. Large polymers are calledmacromolecules.

Monomers link to form polymers through a chemical reactioncalled a condensation reaction. In the condensation reaction shown


C HC

CC

H

C

CC

H

H

H

H

HC

CH

H

H

C OH

H

H

Carbon can form (a) double or even (b) triple bonds to satisfy its need foreight electrons in its outermost energylevel. Organic molecules can have manydifferent shapes and patterns of bond-ing. Organic molecules can also havemany different functional groups, whichinfluence the properties of the moleculethey are attached to. Notice thehydroxyl, —OH, group on this model ofthe alcohol ethanol (c).

FIGURE 3-6

(a) Benzene (b) Acetylene (c) Ethanol

Demonstrating Polarity

Materials disposable gloves; labapron; safety goggles; 3 test tubes;test-tube rack; 6 mL each of cook-ing oil, ethanol, and waterProcedure

1. Put on your disposable gloves,lab apron, and safety goggles.

2. Label the test tubes “A,” “B,”and “C.”

3. In test tube A, put 3 mL ofwater and 3 mL of oil.

4. In test tube B, put 3 mL of oiland 3 mL of ethanol.

5. In test tube C, put 3 mL ofethanol and 3 mL of water.

6. With your thumb and middle finger, flick each test tube tomix the contents, and allow it to sit for 10–15 minutes.

7. Record your observations.Analysis How does this activitydemonstrate polarity of moleculesthat contain the –OH group?

Quick Lab

A polymer is the result of bonding betweenmonomers. The six-sided shape is an organicstructural model of a molecule with a cen-tral carbon ring. The organic structure of a molecule shows the arrangement of carbon atoms in organic molecules.

FIGURE 3-7


in Figure 3-8, two sugar molecules, glucose andfructose, combine to form the sugar sucrose,which is common table sugar. The two sugarmonomers become linked by a C–O–C bridge.In the formation of that bridge, the glucose mol-ecule releases a hydroxide ion, OH�, and thefructose molecule releases a hydrogen ion, H�.The OH� and H� ions that are released in turncombine to produce a water molecule, H2O.

The breakdown of some complex molecules, such as polymers,occurs through a process known as hydrolysis (HIE-DRAH-luh-sis).Hydrolysis is a reversal of a condensation reaction. The addition ofwater to some complex molecules, including polymers, under cer-tain conditions can break the bonds that hold them together. As youcan see in Figure 3-9, in hydrolysis a large molecule breaks apart.

Energy CurrencyLife processes require a constant supply of energy. This energy isavailable to cells in the form of certain compounds that contain alarge amount of energy in their overall structure. One of these com-pounds is adenosine (uh-DEN-uh-SEEN) triphosphate, more commonlyreferred to by its abbreviation, ATP.

Figure 3-9 shows the structure of an ATP molecule. Notice thethree linked phosphate groups, —PO4

�, that are attached to oneanother by covalent bonds. The covalent bond that holds the lastphosphate group to the rest of the molecule is easily broken. Whenthis bond is broken, much more energy is released than was requiredto break the bond. This conversion of energy is used by the cell todrive the chemical reactions that enable an organism to function.

C H A P T E R 354

C

C

O

H

OH

C

OH

H

CH2OH

C

H

H2O

CH2OH

C

HO

H

C

O

H

C

OH

H

C

CH2OH

H

C

H

OH

O

GlucoseSUCROSE

Fructose

Adenosine triphosphate (ATP)

P P P P P

H

P

Adenosine diphosphate (ADP) andinorganic phosphate

H2O

OH

1. What is an organic compound?

2. What property allows carbon compounds to existin a number of forms?

3. Define functional group and give an example.

4. How does a polymer form?

5. How does a polymer break down?

6. CRITICAL THINKING Scientists can determinethe age of a substance using a method that com-pares the amounts of different forms of carbonatoms present in the substance. Is this methodmore useful for organic substances or inorganicsubstances?

SECTION 3-2 REVIEW

The condensation reaction of one glu-cose molecule with one fructose mol-ecule yields sucrose and water. Onewater molecule is produced each timetwo monomers form a covalent bond.

FIGURE 3-8

The hydrolysis of ATP yields adenosinediphosphate and inorganic phosphate.In hydrolysis, a hydrogen ion from a water molecule bonds to one of thenew molecules, and a hydroxide ionbonds to the other new molecule. Mosthydrolysis reactions are exergonic.

FIGURE 3-9


M O L E C U L E S O F L I F EFour main classes of organic compounds are essential to the

life processes of all living things: carbohydrates, lipids,

proteins, and nucleic acids. You will see that although these

compounds are built from carbon, hydrogen, and oxygen, the

atoms occur in different ratios in each class of compound.

Despite their similarities, the different classes of compounds

have different properties.

CARBOHYDRATESCarbohydrates are organic compounds composed of carbon,hydrogen, and oxygen in a ratio of about two hydrogen atoms toone oxygen atom. The number of carbon atoms in a carbohydratevaries. Carbohydrates exist as monosaccharides, disaccharides,or polysaccharides.

MonosaccharidesA monomer of a carbohydrate is called a monosaccharide (MAHN-oh-SAK-uh-RIED). A monosaccharide—or simple sugar—contains carbon,hydrogen, and oxygen in a ratio of 1:2:1. The general formula for amonosaccharide is written as (CH2O)n, where n is any whole numberfrom 3 to 8. For example, a six-carbon monosaccharide (CH2O)6would have the formula C6H12O6. The most common monosaccha-rides are glucose, fructose, and galactose, as shown in Figure 3-10.Glucose is a main source of energy for cells. Fructose is found in fruits and is the sweetest of the monosaccharides. Galactose isfound in milk and is usually combined with glucose or fructose.Notice in Figure 3-10 that glucose, fructose, and galactose have thesame molecular formula, C6H12O6, but their differing structuresdetermine the slightly different properties of the three compounds.Compounds like these sugars, with a single chemical formula but different forms, are called isomers (IE-soh-muhrz).


C

HO

H

C

H

OH

C

OH

H

C

CH2OH

H

C

H

OH

O

Glucose

C

OH C

O

H

OH

C

OH

H

CH2OH

C

H

CH2OH

Fructose

C

H

HO

C

OH

H

C

OH

H

C

CH2OH

H

C

H

OH

O

Galactose

FIGURE 3-10

Although glucose, fructose, and galac-tose have the same chemical formula,their structural differences result in different properties among the threecompounds.

▲

Define monosaccharide,disaccharide, and polysaccharide,

and discuss their significance to organisms.

●

Relate the sequence of amino acids to the structure of proteins.

■

Relate the structure of lipids to their functions.

◆

List two essential functions of nucleic acids.

O B J E C T I V E S

SECTION

3-3

TOPIC: CarbohydratesGO TO: www.scilinks.orgKEYWORD: HM055


Disaccharides and PolysaccharidesIn living things, two monosaccharides can combine in a condensa-tion reaction to form a double sugar, or disaccharide. As you sawin Figure 3-8, sucrose, which is common table sugar, is composedof fructose and glucose. A polysaccharide is a complex moleculecomposed of three or more monosaccharides.

Animals store glucose in the form of the polysaccharide glyco-gen. Glycogen consists of hundreds of glucose molecules strungtogether in a highly branched chain. Much of the glucose thatcomes from food is ultimately stored in your liver and muscles asglycogen and is ready to be used for quick energy.

In plants, glucose molecules are linked in the polysaccharidestarch. Starch molecules have two basic forms—highly branchedchains that are similar to glycogen and long, unbranched chainsthat coil like a telephone cord. The large polysaccharide celluloseis also made by plants. Cellulose, which gives strength and rigidityto plant cells, makes up about 50 percent of wood. In a single cel-lulose molecule, thousands of glucose monomers are linked inlong, straight chains. These chains tend to form hydrogen bondswith each other. The resulting structure is strong and can be bro-ken down by hydrolysis only under certain conditions.

PROTEINSProteins are organic compounds composed mainly of carbon,hydrogen, oxygen, and nitrogen. Like the other macromolecules,proteins are formed from the linkage of monomers. The skin andmuscles of animals are made mostly of proteins, as are many of thecatalysts found in both plants and animals.

Amino AcidsThe 20 different amino acids, the monomer building blocks of pro-teins, share a basic structure. As Figure 3-11 shows, each aminoacid contains a central carbon atom covalently bonded to fourother atoms or functional groups. A single hydrogen atom, high-lighted in blue on the illustration, bonds at one site. A carboxylgroup, —COOH, highlighted in green, bonds at a second site. An amino group, —NH2, highlighted in yellow, bonds at a thirdsite. And a functional group call the R group, highlighted in red,bonds at the fourth site.

The main difference among the different amino acids is found intheir R groups. The R group can be as simple as the single hydro-gen atom of glycine, shown in Figure 3-11a, or it can be more com-plex, such as the R group shown in the model of alanine, shown in Figure 3-11b. The differences among the amino acid R groupsgives different proteins very different shapes. The different shapesallow proteins to perform many different roles in the chemistry ofliving things.

C H A P T E R 356

(a) Glycine

H

CN OHCH

HH O

(b) Alanine

CH3

CN OHCH

HH O

Amino acids differ only in the type ofR group (shown in red) they carry.Glycine (a) has a simpler R group thanalanine (b). The R group may be eitherpolar or nonpolar. Amino acids withpolar R groups can dissolve in water,while those with nonpolar R groupscannot.

FIGURE 3-11


Dipeptides and PolypeptidesFigure 3-12 shows how two amino acids bond to form a dipeptide(die-PEP-tied). In a condensation reaction, two amino acids form acovalent bond, called a peptide bond.

Amino acids can bond to each other one at a time, forming avery long chain called a polypeptide (PAH-lee-PEP-tied). Proteins arecomposed of one or more polypeptides. Some proteins are verylarge molecules, containing hundreds of amino acids. Often theselong proteins are bent and folded upon themselves as a result ofinteractions—such as hydrogen bonding—among individualamino acids. Protein shape can also be influenced by conditionssuch as temperature or the type of solvent in which a protein isdissolved. When you cook an egg, heat changes the shape of pro-teins in the egg white. The firm, opaque result is very different fromthe clear, runny material you began with.

EnzymesRemember from Chapter 2 that enzymes—organic molecules thatact as catalysts—are essential for the functioning of any cell. Mostenzymes are proteins.

Figure 3-13 shows a model of enzyme action. Enzyme reactionsdepend on a physical fit between the enzyme molecule and its sub-strate, the reactant being catalyzed. Notice in Figure 13-3a that theenzyme and substrate have shapes that allow them to fit togetherlike a lock and key. The linkage of the enzyme and substrate causesa slight change in the enzyme’s shape, shown in Figure 13-3b. Thisshape change allows the enzyme to conform to the shape of thesubstrate and probably weakens some chemical bonds in the sub-strate, which is one way that enzymes reduce activation energy.After the reaction, the enzyme releases the products, as shown inFigure 13-3c. Like any catalyst, the enzyme itself is unchanged, soit can be used many times.

An enzyme may fail to work if its environment is changed in someway. For example, change in temperature or pH can cause a changein the shape of the enzyme or the substrate. If this happens, the reac-tion that the enzyme would have catalyzed cannot occur.


H

CN OHCH

HH O

CH3

H2O

Glycine Alanine

CN OHCH

HH O

H

CN CH

HH O

CH3

CN OHC

HH O

The peptide bond that binds aminoacids together results from a conden-sation reaction that produces water.

FIGURE 3-12

(a) In the lock-and-key model ofenzyme action, the enzyme can attachonly to a reactant with a specificshape. (b) The enzyme then flexes toconform to the reactant’s shape. (c) Theenzyme is unchanged by the reaction itparticipates in and is released to beused again.

FIGURE 3-13

Substrate Products

Enzyme(a) (b) (c)


LIPIDSLipids are large, nonpolar organic molecules that do not dissolvein water. Lipid molecules have a higher ratio of carbon and hydro-gen atoms to oxygen atoms than carbohydrates have. Lipids storeenergy efficiently. Lipid molecules have large numbers of carbon-hydrogen bonds, which store more energy than the carbon-oxygenbonds common in other organic compounds.

Fatty AcidsFatty acids are unbranched carbon chains that make up mostlipids. The model in Figure 3-14 shows that a fatty acid contains along, straight carbon chain (from 12 to 28 carbons) with a carboxylgroup, —COOH, attached at one end.

The two ends of the fatty-acid molecule have very differentproperties. The carboxyl end of the fatty-acid molecule is polarand is thus attracted to water molecules. Because of this attrac-tion, the carboxyl end of the fatty-acid molecule is said to behydrophilic (HIE-droh-FIL-ik), which means “water loving.” In contrast,the hydrocarbon end of the fatty-acid molecule is nonpolar. Thisend tends not to interact with water molecules and is said to behydrophobic (HIE-droh-FOH-bik), or “water fearing.”

In saturated fatty acids, like palmitic acid, which is pictured inFigure 3-14, each carbon atom is covalently bonded to four atoms.The carbon atoms are in effect full, or “saturated.” In contrast, youcan see from the structural formula of a molecule of linoleic acid,shown in Figure 3-14, that the carbon atoms are not bonded to themaximum number of atoms that they can bond to. Instead, theyhave formed double bonds within the carbon chain. This type offatty acid is said to be unsaturated.

Complex LipidsLipids are divided into categories according to their structure.Three classes of lipids important to living things contain fattyacids: triglycerides, phospholipids, and waxes. A triglyceride (trie-GLIS-uh-RIED) is composed of three molecules of fatty acid joined toone molecule of the alcohol glycerol. Saturated triglycerides arecomposed of saturated fatty acids. They typically have high melt-ing points and tend to be solid at room temperature. Commondietary saturated triglycerides include shortening and animal fats.In contrast, unsaturated triglycerides are composed of unsaturatedfatty acids and are usually liquid at room temperature. Unsaturatedtriglycerides are found primarily in plant seeds and fruits, wherethey serve as an energy and carbon source for sprouting plants.

Phospholipids have two, rather than three, fatty acids joinedby a molecule of glycerol. As shown in Figure 3-15, the cell mem-brane is composed of two layers of phospholipids, which arereferred to as the lipid bilayer. The inability of lipids to dissolve inwater allows the membrane to form a barrier between the inside

C H A P T E R 358

Phospholipids

Hydrophilic“head”

Phospholipids

The lipid bilayer of a cell membrane ismade from a double row of phospho-lipids, arranged with their hydropho-bic “tails” facing each other. In thisillustration of phospholipid structure,the head represents polar carboxylheads of two fatty acids.

FIGURE 3-15

Water

C HH

CO OH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C HH

C

H

Palmitic acid

HH

C HH

Linoleic acid

C HH

C HH

C HH

C HH

CO OH

C HH

C HH

C HH

C H

C H

C H

C HH

C HH

C H

C HH

C HH

C

H

HH

C HH

Fatty acids have a polar carboxylhead, purple, and a nonpolarhydrocarbon tail, green.

FIGURE 3-14

Hydrophobic“tail”


and outside of the cell. This bilayer arrangement of molecules produces a stable and effective barrier for a cell.

A wax is a type of structural lipid. A wax molecule consists of a long fatty-acid chain joined to a long alcohol chain. Waxes arehighly waterproof, and in plants, wax forms a protective coating on the outer surfaces. Wax also forms protective layers in animals.For example, earwax helps prevent microorganisms from enteringthe middle ear.

SteroidsUnlike most other lipids, which are composed of fatty acids, steroidmolecules are composed of four fused carbon rings with various func-tional groups attached to them. Many animal hormones, such as themale hormone testosterone, are steroid compounds. One of the mostfamiliar steroids in humans is cholesterol. Cholesterol is needed bythe body for nerve cells and other cells to function normally.

NUCLEIC ACIDSNucleic acids are very large and complex organic molecules thatstore important information in the cell. Just as computers use abinary system of zeros and ones to store information, nucleic acidsuse a system of four compounds to store hereditary information. Asequence of the four compounds arranged in a certain order acts as a code for the genetic instructions of the cell.

Deoxyribonucleic acid, or DNA, contains information that isessential for almost all cell activities, including cell division.Ribonucleic (RIE-boh-noo-KLEE-ik) acid, or RNA, stores and transfersinformation that is essential for the manufacturing of proteins.Both DNA and RNA are polymers, composed of thousands of linked monomers called nucleotides (NOO-klee-uh-TIEDS). As shown inFigure 3-16, each nucleotide is made of three main components: aphosphate group, a five-carbon sugar, and a ring-shaped nitrogenbase. You will learn more about these important compounds inChapter 10.


1. Define monosaccharide, disaccharide, andpolysaccharide.

2. Describe the structure of amino acids and proteins.

3. Explain the relationship between an enzyme andits substrate.

4. How do the two ends of a fatty acid differ?

5. Name the two types of nucleic acids, anddescribe their functions.

6. CRITICAL THINKING High temperatures canweaken bonds between different parts of a protein molecule, thus changing its shape. Howmight this change alter the effectiveness of an enzyme?

SECTION 3-3 REVIEW

A nucleotide consists of a phosphategroup, a five-carbon sugar, and a ring-shaped nitrogen base. DNA and RNAare very large molecules formed fromthousands of nucleotides strungtogether in a chain.

FIGURE 3-16

Phosphate group Nitrogen-containingbase

Sugar(deoxyribose)

TOPIC: SteroidsGO TO: www.scilinks.orgKEYWORD: HM059

C H A P T E R 360

nsulin is a hormone secreted bycells within the pancreas. It is

essential in regulating the metabo-lism of carbohydrates and fats inthe body. People with the disorderdiabetes mellitus do not produceenough insulin. Some diabetespatients must take insulin injectionsto maintain normal metabolism.

In 1943, the British biochemistFrederick Sanger set out to analyzethe insulin molecule. He was inter-ested in proteins, and he choseinsulin as the subject of his researchmainly because, as he put it, “It wasthe only protein you could buy inpure form over the counter.” Heknew that an understanding of thestructure of insulin could haveimportant implications for medical

practice. He spent the next 12 yearsstudying the structure of insulin.

Biochemists already knew thatproteins consist of combinations of20 different amino acids linkedtogether in chains. They also knewhow to calculate the proportion ofeach amino acid in a given protein.What they did not know was theorder in which the amino acids are linked in a specific protein.They correctly believed that thesequence of amino acids in a pro-tein is crucial to the protein’s function. Sanger’s goal was todetermine the amino acid sequenceof insulin, and to do so, he had todevelop new laboratory techniques.

Sanger began with a strategyfamiliar to chemists. He broke the

insulin molecule into pieces. Aftersplitting the chains of amino acidsinto short fragments, Sanger cameto understand how they fittogether. In his quest to determinethe structure of insulin, he deviseda new way to label the ends of aprotein fragment.

Sanger learned that insulin ismade up of two linked chains, onecontaining 30 amino acids and theother containing 21. He looked forfragments with overlappingsequences, which helped him dis-cover how each chain was puttogether. By 1952, he had learnedthe amino acid sequences of bothchains, but he still needed to under-stand how the two chains werelinked to make up one insulin mol-ecule. Three years later, Sangerreached his goal of identifying themolecular structure of insulin.

Sanger’s work with insulinestablished him as the leader in his field, and in 1958 he received a Nobel Prize in chemistry. Bydemonstrating that each proteinhas a unique structure and likewisea unique sequence of amino acids,Sanger paved the way for thedevelopment of a technique thatenabled the synthesis of insulin inthe laboratory. In 1980, he receiveda second Nobel Prize in chemistryfor his work in developing tech-niques for determining thesequence of nucleotides in mol-ecules of DNA and RNA. Sanger isone of only four people to ever beawarded two Nobel Prizes.

I

The Structure of Insulin

Frederick Sanger was the first scientist to determine the sequence of amino acids in a protein.



CHAPTER 3 REVIEW

adhesion (51)capillarity (51)

cohesion (51) hydrogen bond (50) polar (50)

■ Water is a polar molecule in which the elec-trons are unevenly shared between thehydrogen and oxygen atoms.

■ Because of its polar nature, water is effec-tive in dissolving other substances to formsolutions.

■ Hydrogen bonding is responsible for thecohesion and capillarity that water mol-ecules display.

■ Water can absorb a large amount of thermalenergy before its own temperature beginsto rise.

SUMMARY/VOCABULARY

Vocabulary

3-1

adenosine triphosphate (ATP)(54)

alcohol (53)

condensation reaction (53)functional group (53)hydrolysis (54)

macromolecule (53)monomer (53)

organic compound (52)polymer (53)

■ An organic compound contains carbon thatis covalently bonded both to other carbonatoms and often to atoms of other ele-ments, including oxygen, hydrogen, andnitrogen.

■ A carbon atom forms four covalent bondswith other atoms. Carbon atoms can bondwith one another to form straight chains,branched chains, or rings.

■ Simple molecules, known as monomers,

bond to one another to form complex mol-ecules called polymers. Monomers arejoined to form polymers in a condensationreaction. Polymers are broken down intomonomers during hydrolysis.

■ An alcohol is an organic compound thatcontains a hydroxyl group, OH�, attachedto one of its carbon atoms.

■ Adenosine triphosphate (ATP) makesenergy available to a cell.

Vocabulary

3-2

■ A carbohydrate is an organic compoundcomposed of carbon, hydrogen, and oxy-gen atoms in a ratio of about two hydrogenatoms to one oxygen atom. A monomer ofa carbohydrate is called a monosaccharide.

■ A double sugar is called a disaccharide. Acomplex sugar made of many monosaccha-rides is called a polysaccharide.

■ A protein is an organic molecule that isformed from amino acids. An amino acidconsists of a central carbon atom to whichfour functional groups are attached.

■ Amino acids are joined by peptide bonds. A longchain of amino acids is called a polypeptide.

■ Enzymes are catalysts that act in living

things. Enzyme action can be explained bythe lock-and-key model. Most enzymes areproteins.

■ Most lipids contain fatty acids, organic mol-ecules that have a hydrophilic end and ahydrophobic end.

■ Unsaturated lipids have one or more pairsof carbon atoms joined by double bonds.Saturated lipids have no double bondsbetween their carbon atoms.

■ Lipids store more energy than the othertypes of organic molecules.

■ Nucleic acids are organic molecules thatstore genetic information in the cell.

3-3

amino acid (56)carbohydrate (55)dipeptide (57)disaccharide (56)fatty acid (58)hydrophilic (58)

hydrophobic (58)isomer (55)lipid (58)monosaccharide (55)nucleic acid (59)nucleotide (59)

peptide bond (57)phospholipid (58)polypeptide (57)polysaccharide (56)protein (56)ribonucleic acid (RNA) (59)

steroid (59)substrate (57)triglyceride (58)wax (59)

Vocabulary


CHAPTER 3 REVIEW

C H A P T E R 362

REVIEW

Vocabulary1. Explain the relationship between amino

acids, peptide bonds, and proteins.2. How are the structures of monosaccharides,

disaccharides, and polysaccharides relatedto each other?

3. Explain the relationship between a polypep-tide and a peptide bond.

4. What is the difference between a hydrophilicsubstance and a hydrophobic substance?

5. Why is the structure of the cell membranereferred to as a bilayer?

Multiple Choice6. Water helps keep the temperature of living

things (a) high (b) low (c) below the freezingpoint (d) stable.

7. The distinguishing feature of a molecule of apolar compound is its (a) even distribution ofelectrical charge (b) uneven distribution ofelectrical charge (c) even temperature (d)uneven temperature.

8. The element that readily bonds to itself,forming long chains and rings, is (a) hydro-gen (b) nitrogen (c) carbon (d) oxygen.

9. Plants store glucose in (a) a polysaccharidecalled starch (b) long proteins (c) complexlipid molecules called triglycerides (d) sim-ple sugar molecules.

10. A very strong structural molecule in plantsthat is formed by hydrogen bonding betweenchains of glucose molecules is (a) starch (b) wax (c) cellulose (d) glycogen.

11. When two amino acids bond, (a) water istaken in by the product (b) hydrolysis occurs(c) a dipeptide is formed through a condensa-tion reaction (d) a triglyceride is formed.

12. Lipids are distinguished from other organicmolecules because they (a) contain carbon,hydrogen, and oxygen in a ratio of 1:2:1 (b) do not dissolve in water (c) dissolve eas-ily in water (d) form large protein molecules.

13. Steroids differ from other lipid polymers inthat steroids (a) do not occur in varied sub-stances (b) are not hydrophilic (c) are nothydrophobic (d) are not composed of fatty-acid monomers.

14. Most enzymes are (a) lipids (b) phospho-lipids (c) proteins (d) carbohydrates.

15. A compound that stores hereditary informa-tion is (a) ATP (b) alcohol (c) DNA (d) protein.

Short Answer16. Label the parts of the nucleotide below.

17. What are isomers?18. What properties do alcohols and water share?19. Compare and contrast a condensation reac-

tion with a hydrolysis reaction.20. Use a diagram to show how enzymes work.21. How does the carboxyl end of the fatty-acid

molecule differ from the hydrocarbon end ofthe molecule?

22. Compare and contrast the structures oftriglycerides, phospholipids, and steroids.Which type of lipid is structurally unlike theother two?

23. What role does the compound ATP play incellular activities?

24. What is an important characteristic of waxes,and why is this valuable to living things?

25. What structural role do phospholipids playin cells?

B

C

A


CRITICAL THINKING

1. Cells contain mostly water. What would hap-pen to the stability of an organism’s internaltemperature with respect to environmentaltemperature changes if cells containedmostly oil, which does not have extensivehydrogen bonding?

2. The surface tension of water at room temper-ature is so great that you can actually “float”a small needle on the surface of water. (Theneedle doesn’t truly float—it is denser thanwater. It is held in place by the force ofhydrogen bonding between water moleculeslying below and around the needle.) If thewater were heated, what would happen tothe needle and why?

3. Starch easily dissolves in water. Cellulosedoes not. Both substances, however, consistof chains of glucose molecules. What struc-tural difference between starch and celluloseaccounts for this different behavior in water?

4. Triglycerides in animals’ bodies are usuallysolid fats, and those in plants are usuallyoils. However, many animals living in theArctic and Antarctic have a greater numberof triglycerides that are oils than do otheranimals. What advantage would the storageof body fat as oil instead of solid fat be toanimals that live in freezing climates?

5. The specific heat of a substance is theamount of heat that must be added to 1 g ofthe substance to raise its temperature 1°C.Specific heat is measured in calories. Use thetable to answer the following questions. a. What substance can absorb the greatest

amount of heat before its own temperaturerises?

b. What do the two substances with the low-est specific heat values have in common?

c. What property of ethanol might accountfor its relatively high specific heat?

d. Which would be a better conductor ofheat: iron or glass?

e. What practical use would a substance witha very high specific heat have?

1. Read “Alien Haven” in New Scientist,September 18, 1999, on page 32, and answerthe following questions: What are the“Goldilocks criteria” for a planet to be idealfor life? What is the “habitable zone,” wherea planet that can support life is most likelyto be found?

2. Cut fibrous meat into four 1 in. cubes.Sprinkle three of the cubes with equalamounts of meat tenderizer, which containsa protein-splitting enzyme called papain.Place one cube in the refrigerator, leave oneat room temperature, and place the other in

an incubator at 32°C. For the fourth cube,place the same amount of meat tenderizerand a few tablespoons of water in a con-tainer, and boil the mixture for three min-utes. (Do not allow the mixture to boil dry.Add water, tablespoon by tablespoon, asneeded.) Pour the boiled mixture on themeat. After three hours, observe the textureof all four meat cubes. What do you con-clude about the effect of temperature on theenzyme in meat tenderizer? Express yourresults as a graph showing temperature andthe apparent level of activity of the enzyme.

EXTENSION

CHAPTER 3 REVIEW


Specific Heats of Common Substances

Substance

Lead

Iron

Glass

Ethanol

Water

Liquid ammonia

Specific heat (cal)

0.03

0.10

0.20

0.60

1.00

1.23


Identifying Organic Compounds in Foods

CHAPTER 3 INVESTIGATION

C H A P T E R 364

OBJECTIVES

■ Determine whether specific nutrients are present in a solution of unknown composition.

■ Perform chemical tests using substances called indicators.

PROCESS SKILLS

■ experimenting■ observing■ measuring

MATERIALS

■ lab apron ■ glucose solution■ safety goggles ■ unknown solution■ disposable gloves ■ distilled water■ 1 L beaker ■ 9 glass stirring rods■ hot plate ■ tongs or test-tube holder■ 9 test tubes ■ test-tube rack■ labeling tape ■ albumin solution■ marker ■ sodium hydroxide solution■ 10 mL graduated cylinder ■ copper sulfate solution■ Benedict’s solution ■ vegetable oil■ 9 dropping pipets ■ Sudan III solution

Background

1. Carbohydrates, proteins, and lipids are nutrients thatare essential to all living things. Some foods, such astable sugar, contain only one of these nutrients. Mostfoods, however, contain mixtures of proteins, carbohy-drates, and lipids. You can confirm this by reading theinformation in the “Nutrition Facts” box found on anyfood label.

2. In this investigation, you will use chemical substances,called indicators, to identify the presence of specificnutrients in an unknown solution. By comparing thecolor change an indicator produces in the unknownfood sample with the change it produces in a sampleof known composition, you can determine whetherspecific organic compounds are present in theunknown sample.

3. Benedict’s solution is used to determine the presenceof monosaccharides, such as glucose. A mixture ofsodium hydroxide and copper sulfate determine thepresence of some proteins (this procedure is called thebiuret test). Sudan III is used to determine the pres-ence of lipids.

Procedure

CAUTION Put on a lab apron,safety goggles, and gloves. In

this lab, you will be working with chemicals that canharm your skin and eyes or stain your skin and cloth-ing. If you get a chemical on your skin or clothing,wash it off at the sink while calling to your teacher.If you get a chemical in your eyes, immediately flushit out at the eyewash station while calling to yourteacher. As you perform each test, record your data inyour lab report, organized in a table like the one on thenext page.

Test 1

1. CAUTION Do not touch the hotplate. Use tongs to move

heated objects. Turn off the hot plate when notin use. Do not plug in or unplug the hot platewith wet hands. Make a water bath by filling a 1 Lbeaker half full with water. Then put the beaker on ahot plate and bring the water to a boil.

2. While you wait for the water to boil, label one test tube“1-glucose,” label the second test tube “1-unknown,”and label the third test tube ”1-water.” Using the grad-uated cylinder, measure 5 mL of Benedict’s solution andadd it to the “1-glucose” test tube. Repeat the proce-dure, adding 5 mL of Benedict’s solution each to the “1-unknown” test tube and “1-water” test tube.

3. Using a dropping pipet or eyedropper, add 10 drops ofglucose solution to the “1-glucose” test tube. Using asecond dropping pipet, add 10 drops of the unknownsolution to the “1-unknown” test tube. Using a thirddropping pipet, add 10 drops of distilled water to the


“1-water” test tube. Mix the contents of each testtube with a clean stirring rod. (It is important not tocontaminate test solutions by using the samedropping pipet or stirring rod in more than onesolution. Use a different dropping pipet and stir-ring rod for each of the test solutions.)

4. When the water boils, use tongs to place the test tubesin the water bath. Boil the test tubes for 1 to 2 minutes.

5. CAUTION Do not touch the test tubes withyour hands. They will be very hot. Use tongs

to remove the test tubes from the water bath andplace them in the test-tube rack. As the test tubescool, an orange or red precipitate will form if largeamounts of glucose are present. If small amounts ofglucose are present, a yellow or green precipitate willform. Record your results in your data table.

Test 2

6. Label one clean test tube “2-albumin,” label a second test tube “2-unknown,” and label

a third test tube “2-water.” Using a dropping pipet,add 40 drops of albumin solution to the “2-albumin”test tube. Using a second dropping pipet, add 40drops of unknown solution to the “2-unknown” testtube. Using a third dropping pipet, add 40 drops ofwater to the “2-water” test tube.

7. Add 40 drops of sodium hydroxide solution to each ofthe three test tubes. Mix the contents of each testtube with a clean stirring rod.

8. Add a few drops of copper sulfate solution, one dropat a time, to the “2-albumin” test tube. Stir the solu-tion with a clean stirring rod after each drop. Note thenumber of drops required to cause the color of thesolution in the test tube to change. Then add the samenumber of drops of copper sulfate solution to the “2-unknown” and “2-water” test tubes.

9. Record your results in your data table.

Test 3

10. Label one clean test tube “3-vegetable oil,” label asecond test tube “3-unknown,” and label a third testtube “3-water.” Using a dropping pipet, add 5 dropsof vegetable oil to the “3-vegetable oil” test tube.Using a second dropping pipet, add 5 drops of theunknown solution to the “3-unknown” test tube.Using a third dropping pipet, add 5 drops of water tothe “3-water” test tube.

11. CAUTION Sudan III solution will stain yourskin and clothing. Promptly wash off spills

to minimize staining. Do not use Sudan III solu-tion in the same room with an open flame. Usinga clean dropping pipet, add 3 drops of Sudan III solu-tion to each test tube. Mix the contents of each testtube with a clean stirring rod.

12. Record you results in your data table.13. Clean up your materials and wash your

hands before leaving the lab.

Analysis and Conclusions

1. Based on the results you recorded in your data table,identify the nutrient or nutrients in the unknown solution.

2. What are the experimental controls in this investigation?3. Explain how you were able to use the color changes

of different indicators to determine the presence ofspecific nutrients in the unknown substance.

4. List four potential sources of error in this investigation.

Further Inquiry

Is there a kind of macromolecule that the tests in this lab didnot test for? If so, list the kind of macromolecules not testedfor, and give one reason why they were not tested for.

TABLE A IDENTIFICATION OF SPECIFIC NUTRIENTS BY CHEMICAL INDICATORS

Nutrient category(protein, lipid, etc.)

Nutrient in test solution

Result for known sample

Result for unknown sample

Result for distilled waterTest

1

2

3



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